Determine the x values that cause the function to be (a) zero, (b) undefined, (c) positive, and (d) negative.
-$$f ( x ) = \frac { ( x - 2 ) } { ( 2 x + 1 ) ( x - 3 ) }$$
A) (a) $ \left\{2,3,-\frac{1}{2}\right\} $ (b) $ \varnothing $, (c) $ \left(-\infty,-\frac{1}{2}\right) \cup(2,3) $, (d) $ \left(-\frac{1}{2}, 2\right) \cup(3, \infty) $
B) (a) $ \{3\} $, (b) $ \left\{3,-\frac{1}{2}\right\} $, (c) $ \left(-\infty,-\frac{1}{2}\right) \cup(2,3),(\mathrm{d})\left(-\frac{1}{2}, 2\right) \cup(3, \infty) $
C) (a) $ \{2,3\} $, (b) $ \left\{-\frac{1}{2}\right\} $ (c) $ \left\{-\frac{1}{2}, 2\right) \cup(3, \infty),(\mathrm{d})\left(-\infty,-\frac{1}{2}\right) \cup(2,3) $
D) (a) $ \{2\} $, (b) $ \left\{3,-\frac{1}{2}\right\} $, (c) $ \left\{-\frac{1}{2}, 2\right) \cup(3, \infty),(d)\left(-\infty,-\frac{1}{2}\right) \cup(2,3) $

Question

Quizplus · Accepted Answer

(a) The function is zero when the numerator is zero, which is at $x = 2$. (b) The function is undefined when the denominator is zero, which occurs at $x = 3$ and $x = -\frac{1}{2}$. (c) The function is positive when the product of the signs of the numerator and denominator is positive, which occurs in the intervals $(- \frac{1}{2}, 2)$ and $(3, \infty)$. (d) The function is negative when the product of the signs of the numerator and denominator is negative, which occurs in the intervals $(- \infty, -\frac{1}{2})$ and $(2, 3)$.

Determine the X Values That Cause the Function to Be f(x)=(x−2)(2x+1)(x−3)f ( x ) = \frac { ( x - 2 ) } { ( 2 x + 1 ) ( x - 3 ) }f(x)=(2x+1)(x−3)(x−2)​

Determine the X Values That Cause the Function to Be $f ( x ) = \frac { ( x - 2 ) } { ( 2 x + 1 ) ( x - 3 ) }$